How to remove even numbers in List using Prolog

Question

I need to remove all even numbers in first list and save the rest to second list. My first non-working approach was:

remove_even([],[]).
remove_even([H1|T1],         


        

Answer 1

If your Prolog system offers clpfd, then this answer is for you. (If not, please read on, too!)

By using meta-predicate tfilter/3 and the reified predicate zodd_t/2, simply write:

?- tfilter(zodd_t,[1,2,3,4],Zs). % remove even integers = keep odd integers
Zs = [1,3].

As we only use monotone code here, we get sound answers for more general queries, too:

?- tfilter(zodd_t,[A,B,C],Zs).
Zs = [     ], A mod 2 #= 0, B mod 2 #= 0, C mod 2 #= 0 ;
Zs = [    C], A mod 2 #= 0, B mod 2 #= 0, C mod 2 #= 1 ;
Zs = [  B  ], A mod 2 #= 0, B mod 2 #= 1, C mod 2 #= 0 ;
Zs = [  B,C], A mod 2 #= 0, B mod 2 #= 1, C mod 2 #= 1 ;
Zs = [A    ], A mod 2 #= 1, B mod 2 #= 0, C mod 2 #= 0 ;
Zs = [A,  C], A mod 2 #= 1, B mod 2 #= 0, C mod 2 #= 1 ;
Zs = [A,B  ], A mod 2 #= 1, B mod 2 #= 1, C mod 2 #= 0 ;
Zs = [A,B,C], A mod 2 #= 1, B mod 2 #= 1, C mod 2 #= 1.

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We can use above code even without clpfd, simply by using different reified predicates: eveninteger_t/2 → zeven_t/2, and oddinteger_t/2 → zodd_t/2.

Sample query:

?- tfilter(oddinteger_t,[1,2,3,4,5,6,7],Xs).
Xs = [1,3,5,7].

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Edit 2015-06-06

Let's try the following variation!

Instead of using zodd_t/2 directly, rather use zeven_t/2 and not_t/3:

truth_negated(true,false).
truth_negated(false,true).

not_t(Goal,Param,Truth1) :-       % meta-predicate, negate truth value
   call(Goal,Param,Truth0),
   truth_negated(Truth0,Truth1).

Let's see it if works with a head-to-head comparison:

?- tfilter(      zodd_t  ,[1,2,3,4],Zs).
Zs = [1,3].

?- tfilter(not_t(zeven_t),[1,2,3,4],Zs).
Zs = [1,3].